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## MAXIMUM FLOW

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**MAXIMUM FLOW**Max-Flow Min-Cut Theorem (Ford Fukerson’s Algorithm)**What is Network Flow ?**Flow network is a directed graph G=(V,E) such that each edge has a non-negative capacity c(u,v)≥0. • Two distinguished vertices exist in G namely : • Source (denoted by s) : In-degree of this vertex is 0. • Sink (denoted by t) : Out-degree of this vertex is 0. Flow in a network is an integer-valued function f defined On the edges of G satisfying 0≤f(u,v)≤c(u,v), for every Edge (u,v) in E.**What is Network Flow ?**• Each edge (u,v) has a non-negative capacity c(u,v). • If (u,v) is not in E assume c(u,v)=0. • We have source s and sink t. • Assume that every vertex v in V is on some path from s to t. Following is an illustration of a network flow: c(s,v1)=16 c(v1,s)=0 c(v2,s)=0 …**Conditions for Network Flow**For each edge (u,v) in E, the flow f(u,v) is a real valued function that must satisfy following 3 conditions : • Capacity Constraint : u,v V, f(u,v) c(u,v) • Skew Symmetry : u,v V, f(u,v)= -f(v,u) • Flow Conservation: u V – {s,t} f(s,v)=0 • vV Skew symmetry condition implies that f(u,u)=0.**The Value of a Flow.**The value of a flow is given by : The flow into the node is same as flow going out from the node and thus the flow is conserved. Also the total amount of flow from source s = total amount of flow into the sink t.**1**8, 8 10 9 , s t 1, 1 2 6, 6 10, 7 Example of a flow Capacity Flow Table illustrating Flows and Capacity across different edges of graph above: fs,1 = 9 , cs,1 = 10 (Valid flow since 10 > 9) fs,2 = 6 , cs,2 = 6 (Valid flow since 6 ≥ 6) f1,2 = 1 , c1,2 = 1 (Valid flow since 1 ≥ 1) f1,t = 8 , c1,t = 8 (Valid flow since 8 ≥ 8) f2,t = 7 , c2,t = 10 (Valid flow since 10 > 7) The flow across nodes 1 and 2 are also conserved as flow into them = flow out.**In simple terms maximize**the s to t flow, while ensuring that the flow is feasible. The Maximum Flow Problem Given a Graph G (V,E) such that: xi,j = flow on edge (i,j) ui,j= capacity of edge (i,j) s = source node t = sink node Maximize v Subject To Σjxij - Σjxji = 0 for each i ≠s,t Σjxsj = v 0 ≤ xij≤ uij for all (i,j)E.**Cuts of Flow Networks**A Cut in a network is a partition of V into S and T (T=V-S) such that s (source) is in S and t (target) is in T. 2 5 9 Cut 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 4 7 30**Capacity of Cut (S,T)**2 5 9 Cut 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 Capacity = 30 4 7 30**Min Cut**Min s-t cut (Also called as a Min Cut) is a cut of minimum capacity 2 5 9 Min s-t Cut 10 15 15 10 4 5 s 3 6 t 8 10 15 4 6 10 15 Capacity = 28 4 7 30**Flow of Min Cut (Weak Duality)**Let f be the flow and let (S,T) be a cut. Then | f | ≤ CAP(S,T). In maximum flow, minimum cut problems forward edges are full or saturated and the backward edges are empty because of the maximum flow. Thus maximum flow is equal to capacity of cut. This is referred to as weak duality. Proof : S T 0 8 t s 7**Methods**Max-Flow Min-Cut Theorem • The Ford-Fulkerson Method • The Preflow-Push Method**1**1 0 8, 8 1 10 9 , s s t t 9 8 0 1, 1 1 7 6 2 2 0 6, 6 10, 7 3 Residual Network The Ford-Fulkerson Method • Try to improve the flow, until we reach the maximum value of the flow • The residual capacity of the network with a flow f is given by: The residual capacity (rc) of an edge (i,j) equals c(i,j) – f(i,j) when (i,j) is a forward edge, and equals f(i,j) when (i,j) is a backward edge. Moreover the residual capacity of an edge is always non-negative. Original Network**The Ford-Fulkerson Method**Begin x := 0; // x is the flow. create the residual network G(x); while there is some directed path from s to t in G(x) do begin let P be a path from s to t in G(x); Δ:= δ(P); send Δunits of flow along P; update the r's; end end {the flow x is now maximum}. Click To See Ford Fulkerson’s Algorithm In Action (Animation)**Augmenting Paths ( A Useful Concept )**Definition: An augmenting path p is a simple path from s to t on a residual network that is an alternating sequence of vertices and edges of the form s,e1,v1,e2,v2,...,ek,t in which no vertex is repeated and no forward edge is saturated and no backward edge is free. Characteristics of augmenting paths: • We can put more flow from s to t through p. • The edges of residual network are the edges on which residual capacity • is positive. • We call the maximum capacity by which we can increase the flow on p • the residual capacity of p.**The Ford-Fulkerson’s Algorithm**AUGMENT(f,P) b bottleneck(P) FOREACH e P IF (e E) // backwardsarc f(e) f(e) + b ELSE // forward arc f(eR) f(e) - b RETURN f FORDFULKERSON(G,E,s,t) FOREACH e E f(e) 0 Gf residual graph WHILE (there exists augmenting path P) f augment(f, P) update Gf ENDWHILE RETURN f Click To See Ford Fulkerson’s Algorithm In Action (Animation)**Proof of correctness of the algorithm**Lemma: At each iteration all residual capacities are integers. Proof: It’s true at the beginning. Assume it’s true after the first k-1 augmentations, and consider augmentation k along path P. The residual capacity Δ of P is the smallest residual capacity on P, which is integral. After updating, we modify the residual capacities by 0 or Δ, and thus residual capacities stay integers. Theorem: Ford-Fulkerson’s algorithm is finite Proof: The capacity of each augmenting path is atleast 1. The augmentation reduces the residual capacity of some edge (s,j) and doesn’t increase the residual capacity for some edge (s,i) for any i. So the sum of residual capacities of edges out of s keeps decr- easing, and is bounded below 0. Number of augmentations is O(nC) where C is the largest of the capacity in the network.**When is the flow optimal ?**• A flow f is maximum flow in G if : • The residual network Gf contains no more augmented paths. • | f | = c(S,T) for some cut (S,T) (a min-cut) Proof: • Suppose there is an augmenting path in Gf then it implies that • the flow f is not maximum, because there is a path through which • more data can flow. Thus if flow f is maximum then residual n/w • Gf will have no more augmented paths. • (2) Let v=Fx(S,T) be the flow from s to t. By assumption v=CAP(S,T) • By Weak duality, the maximum flow is at most CAP(S,T). Thus the • flow is maximum.**15.082 and 6.855J (MIT OCW)**The Ford-Fulkerson Augmenting Path Algorithm for the Maximum Flow Problem**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 2 2 s 4 t 3 2 1 3 This is the original network, and the original residual network.**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 2 2 s 4 t 3 2 1 3 Find any s-t path in G(x)**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 2 1 2 s 4 t 1 2 3 2 1 1 1 3 Determine the capacity D of the path. Send D units of flow in the path.Update residual capacities.**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 2 1 2 s 4 t 1 2 3 2 1 1 1 3 Find any s-t path**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 2 1 1 1 s 4 t 1 1 2 3 2 2 1 1 1 1 1 3 Determine the capacity D of the path. Send D units of flow in the path.Update residual capacities.**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 1 2 1 1 s 4 t 1 1 2 3 2 2 1 1 1 1 1 3 Find any s-t path**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 1 1 1 s 4 t 2 1 2 1 2 3 1 1 1 1 1 3 Determine the capacity D of the path. Send D units of flow in the path.Update residual capacities.**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 1 2 1 1 s 4 t 1 2 2 1 2 3 1 1 1 1 1 3 Find any s-t path**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 1 2 1 1 s 4 t 1 2 2 1 1 2 1 1 1 2 1 2 1 3 Determine the capacity D of the path. Send D units of flow in the path.Update residual capacities.**Ford-Fulkerson Max Flow**4 2 5 1 3 1 1 2 1 1 1 s 4 t 2 1 1 2 1 2 1 1 2 1 2 1 3 Find any s-t path**Ford-Fulkerson Max Flow**4 3 2 5 1 1 3 2 1 1 1 1 s 4 t 2 1 2 1 1 2 1 1 2 1 2 1 3 Determine the capacity D of the path. Send D units of flow in the path.Update residual capacities.**Ford-Fulkerson Max Flow**3 4 2 5 1 1 3 2 1 1 1 1 s 4 t 1 2 1 2 1 2 1 1 2 1 2 1 3 There is no s-t path in the residual network. This flow is optimal**Ford-Fulkerson Max Flow**4 3 2 2 5 5 1 1 3 2 1 1 1 1 s s 4 4 t 1 2 2 1 1 2 1 1 2 1 2 1 3 3 These are the nodes that are reachable from node s.**1**2 5 1 1 2 2 s 4 t 2 2 3 Ford-Fulkerson Max Flow Here is the optimal flow